When proving that there is not rational number $m/n$ equal to $\sqrt{2}$, why does $m$ and $n$ must be not both even? I've read this on Rudin's Principles of Mathematical Analysis: > **1.1 Example** We now show that the equation > > $$p^2=2$$ > > is not satisfied by any rational $p$. If there were such a $p$, we could write $p=m/n$ where $m$ and $n$ are integers **that are not both even**. Let us assume this is done, then it implies > > $$m^2=2n^2$$ My doubt is: Why not both even? I was thinking that it has some relation with $m^2=2n^2$ but this premise is given _before_ the enunciation of $m^2=2n^2$. I guess there is some property for rational numbers that has some relation with this, but I'm unaware of. The question may be trivial but I can't figure it out. I guess I understand the rest of the proof quite well, but I'm stuck at this statement.

If they are both even, we can divide the equation by $4$, getting $m'^2=2n'^2$ with $m=2m'$ and $n=2n'$. We can keep doing this until at least one is not even. As each has a finite number of factors of $2$, we can only do it a finite number of times.